a) Đặt \(\left\{ \begin{array}{l}u = 1 - 2x\\dv = {e^x}dx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = - 2dx\\v = {e^x}\end{array} \right.\)
Khi đó \(\int {(1 - 2x){e^x}} dx\)\( = \left( {1 - 2x} \right){e^x} + \int {2{e^x}dx} \) \( = \left( {1 - 2x} \right){e^x} + 2{e^x} + C\)\( = \left( {3 - 2x} \right){e^x} + C\) b) Đặt \(\left\{ \begin{array}{l}u = x\\dv = {e^{ - x}}dx\end{array} \right.\) \( \Rightarrow \left\{ \begin{array}{l}du = dx\\v = - {e^{ - x}}\end{array} \right.\)
Khi đó \(\int {x{e^{ - x}}dx} \)\( = - x{e^{ - x}} + \int {{e^{ - x}}dx} \)\( = - x{e^{ - x}} - {e^{ - x}} + C\)\( = - \left( {1 + x} \right){e^{ - x}} + C\)
c) Đặt \(\left\{ \begin{array}{l}u = \ln \left( {1 - x} \right)\\dv = xdx\end{array} \right.\) \( \Rightarrow \left\{ \begin{array}{l}du = - \dfrac{1}{{1 - x}}dx\\v = \dfrac{{{x^2}}}{2}\end{array} \right.\)
Khi đó \(\int {x\ln (1 - x)dx} \)\( = \dfrac{{{x^2}}}{2}\ln \left( {1 - x} \right) + \int {\dfrac{{{x^2}}}{{2\left( {1 - x} \right)}}dx} \) \( = \dfrac{{{x^2}}}{2}\ln \left( {1 - x} \right) + \dfrac{1}{2}\int {\left( { - 1 - x + \dfrac{1}{{1 - x}}} \right)dx} \)
\( = \dfrac{{{x^2}}}{2}\ln \left( {1 - x} \right) - \dfrac{1}{2}\int {\left( {\left( {1 + x} \right) - \dfrac{1}{{1 - x}}} \right)dx} \) \( = \dfrac{{{x^2}}}{2}\ln \left( {1 - x} \right) - \dfrac{1}{2}.\dfrac{{{{\left( {1 + x} \right)}^2}}}{2} - \dfrac{1}{2}\ln \left( {1 - x} \right) + C\)
\( = \dfrac{{{x^2}}}{2}\ln \left( {1 - x} \right) - \dfrac{1}{2}\ln \left( {1 - x} \right) - \dfrac{1}{4}{\left( {1 + x} \right)^2} + C\).
d) Ta có: \(\int {x{{\sin }^2}xdx} = \int {x.\dfrac{{1 - \cos 2x}}{2}dx} \) \( = \int {\left( {\dfrac{x}{2} - \dfrac{{x\cos 2x}}{2}} \right)dx} \) \( = \dfrac{{{x^2}}}{4} - \dfrac{1}{2}\int {x\cos 2xdx} \)
Đặt \(\left\{ \begin{array}{l}u = x\\dv = \cos 2xdx\end{array} \right.\) \( \Rightarrow \left\{ \begin{array}{l}du = dx\\v = \dfrac{{\sin 2x}}{2}\end{array} \right.\)
Khi đó \(\int {x\cos 2xdx} \)\( = \dfrac{{x\sin 2x}}{2} - \int {\dfrac{{\sin 2xdx}}{2}} \) \( = \dfrac{{x\sin 2x}}{2} + \dfrac{{\cos 2x}}{4} + C\)
Vậy \(\int {x{{\sin }^2}xdx} \)\( = \dfrac{{{x^2}}}{4} - \dfrac{1}{2}\left( {\dfrac{{x\sin 2x}}{2} + \dfrac{{\cos 2x}}{4} + C} \right)\)\( = \dfrac{{{x^2}}}{4} - \dfrac{1}{4}x\sin 2x - \dfrac{1}{8}\cos 2x + D\).
